exercise 11.1
ZIP with MATLAB scripts and note:
exercise 11.1 notes:
On right hand side, [POZAR] table showing Shottky key parameters, the diode of interest Skyworks SMS1546 on pole position of this table.
[ref1] Skyworks SMS1546 key parameters as shown in Skyworks website. Also attached SMS15464 datasheet.To buy a sample dial (+1) 800 777 7334.
f0=10e9 % [Hz]
Cj=.38e-12 % [F] np or metal-n junction capacitance
rs=4 % [ohm] diode contact and spreading resistance
Is=.3e-6 % [A]
Lp=0
Cp=0
Ls=1e-9 % [H] pg527 diode package contacts
Cp=.07e-12 % [F] diode package contacts
I0=[0 20 50]*1e-6
alpha=1/(25*1e-3) % [1/mV]
Rj=1./(alpha*(I0+Is)) % [ohm] non-linear resistor
Gd=1./Rj % diode dynamic conductance
% for V=V0+v0*cos(w0*t)
% I=I0+v0^2/4*Gd1+v0*Gd*cos(w0*t)+v0^2/4*cos(2*w0*t)
% dI_dc=v0^2*Gd1/4
% I_dc variations considering
% I=I0+v0^2/4*Gd1+v0*Gd*cos(w0*t) only
% Pin_RF=v0^2*Gd/2
% beta_i=dI_dc/Pin_RF
% current sensitivity seems constant
Gd1=alpha^2*(I0+Is)
beta_i=Gd1./(2*Gd)
% voltage sensitivity
beta_v=beta_i.*Rj
f0 = 1.000000000000000e+10
Cj = 3.800000000000000e-13
rs = 4
Is = 3.000000000000000e-07
Lp = 0
Cp = 0
Ls = 1.000000000000000e-09
Cp = 7.000000000000001e-14
I0 = 1.0e-04 *
0 0.200000000000000 0.500000000000000
alpha =
40
Rj = 1.0e+04 *
8.333333333333332 0.123152709359606 0.049701789264414
Gd =
0.000012000000000 0.000812000000000 0.002012000000000
Gd1 =
0.000480000000000 0.032480000000000 0.080480000000000
beta_i =
19.999999999999996 19.999999999999996 20.000000000000000
beta_v =
1.0e+06 *
1.666666666666666 0.024630541871921 0.009940357852883
The obtained beta_v does not agree with solutions manual 3rd column in supplied table.
Try 1, to find out whether the provided beta_v in solutions manual is really correct, or where I went wrong from Rj to beta_v through beta_i
Z1=1./(1./(2*pi*f0*Cj)+Rj)+rs
Zin=1./(1./Z1+1./(2*pi*f0*Cp))+2*pi*f0*Ls
Pin=.5*(I0+Is).^2.*Zin
(Gd1./Pin).^.5
syms V_0
V0=zeros(1,numel(Rj))
for k=1:1:numel(Rj)
eq1=Is*(exp(alpha*V_0)-1)/Zin(k)-V_0/(Rj(k)+rs)==0
V0(k)=double(solve(eq1,V_0))
end
Gd1=alpha^2*Is*exp(alpha*V0)
beta_i=Gd1./(2*Gd) % [A/W]
beta_v=beta_i./Rj % [V/W] it's not supposed to be constant
V0 =
0 0 0
Gd1 =
1.0e-03 *
0.480000000000000 0.480000000000000 0.480000000000000
beta_i =
19.999999999999996 0.295566502463054 0.119284294234592
beta_v =
1.0e-03 *
0.240000000000000 0.240000000000000 0.240000000000000
Try 2:
k=1
syms V
eq1=(Is^2*(exp(alpha*V)-1))^2*Zin(k)-(I0(k)+Is)==0
V0=double(solve(eq1,V))
V0=V0(2) % ignore solution with imaginary component
Gd1=alpha^2*Is*exp(alpha*V0)
beta_i=Gd1./(2*Gd)
beta_v=beta_i.*Rj
V0 =
0.511473975117513
Gd1 =
3.685142096437841e+05
beta_i =
1.0e+10 *
1.535475873515767 0.022691761677573 0.009157907794329
beta_v =
1.0e+15 *
1.279563227929806 0.000279455193074 0.000045516440330
Don't know yet how to obtain
beta_v=[8.7 6.4 4.6] % [V/mW]
and perhaps the supplied values should be in [mV/mW] or [V/W] the reason why [V/mW] units supplied not clear.
Voltage sensitivity beta_v is supposed to be constant, or it should me measurable on a scope and show like graph slope in figure 11.5 [POZAR] pg529.
Since the carrier frequency is supplied f0 and all diode model parameters have been made available, one way around, would be, like in example 10.4; generate signals, build a DC + AC diode model, the AC part of the model as shown in [POZAR] pg527, then cherry-pick Taylor terms as done in [POZAR] explanation for the DC part of the diode model ( that in my opinion, the DC model, needs a bit of further development or at least explanation), to precisely avoid reaching a constant Gd1/(2*Gd), and build a test set to obtain Pin and d_Idc .
Instead I am going to explore a diode test set already implemented with simulink and leave this question open regarding beta_v.
Included in .zip for this exercise Click on green Scope and the expected
there's example_power_diode.xls time signals show up:
The diode block has the following parameters And the block tagged Continuous has a button to generate reports
And these are the reports obtained from Continuous
example_power_diode_report2.rep
Simscape Power Systems (Specialized Technology) Report.
generated by powergui,
28-Apr-2020 17:42:16
Model : \example_power_diode.slx.
[1] Steady-State voltages and currents:
States at 0 Hz :
0.0000e+00 A ---> 'Il_R L '
0.0000e+00 V ---> 'Uc_snubber: Diode'
Measurements at 0 Hz :
0.0000e+00 V ---> U_Vload
0.0000e+00 A ---> I_Iload
Sources at 0 Hz :
0.0000e+00 V ---> U_120V 60Hz
Nonlinear elements at 0 Hz :
[2] Initial values of States Variables:
1 'Il' R L = 1.8088e-01 A
2 'Uc' snubber: Diode = -3.8006e+00 V
[3] Machine Load Flow solution:
example_power_diode.net
Unit specified : OMU
rlc matrix:
Node_1 Node_2 Type R(ohms) L(mH) C(uF)/U(V) Branch# Block name
------------------------------------------------------------------------------------
2 0 S 1 1 0 1 R L
3 2 S 20 0 4 2 snubber: Diode
4 5 S 0.001 0 0 3 Ron switch: Diode
5 2 S 1 0 0 4 SPID Diode
Number of nodes: 5
Number of branches: 4
Number of transformers: 0
Number of mutuals (inductive coupling): 0
Number of voltage sources: 2
Number of current sources: 0
Number of switches: 0
Source matrix:
Node1 Node2 U/I Mag. Phase Frequency Type Block name
(0/1) (V/A) (degrees) (Hz)
--------------------------------------------------------------
3 0 0 120 0 60 22 120V 60Hz
3 4 0 0.8 0 0 21 Vf: Diode
Total number of inductances and capacitors: 2
[ref1]
